How do you solve #5r - 1= 2( r - 4) - 6#?

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Answer 1 · 2017-10-25T05:08:06

#r = -13/3 or -4(1/3)#

Bringing r terms to LHS and constants to RHS,
#5r-2r = -8 - 6 +1#
#3r = -13#
#r = -13/3#

Answer 2 · 2017-10-25T05:08:47

#r=-13/3#

#5r-1=2(r-4)-6#

Distribute the parenthesis
#5r-1=2r-8-6#

Isolate the variable #x#
#5r-2r=-8-6+1#

Combine-like terms
#3r=-13#

Solve for #r#
#r=-13/3#